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Electron-Positron Collision Calculations

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data

    An electron [itex]e^{-}[/itex] and a positron [itex]e^{+}[/itex], each with a total energy of exactly 5000 MeV, collide head-on and annihilate to produce a tauon/anti-tauon pair, [itex]\tau^{-}[/itex] and [itex]\tau^{+}[/itex], and no other particles.

    1. For the electron moving before the collision, state or calculate the mass energy of the electron, in MeV.

    2. Show that the [itex]\tau^{-}[/itex] and [itex]\tau^{+}[/itex] move in exactly opposite directions with equal magnitudes of momentum.

    3. Assuming that the [itex]\tau^{-}[/itex] and [itex]\tau^{+}[/itex] have exactly equal masses, what is the total energy of each?

    4. If [itex]\tau^{-}[/itex] and [itex]\tau^{+}[/itex] both have momentum of magnitude 4675 MeV/c, estimate the tauon mass.

    2. Relevant equations

    Within the problem statement and solution attempt.

    3. The attempt at a solution

    1. Is this just 5000 MeV? or do I use:

    [tex]E=mc^{2}\implies m_{e}=\frac{E}{c^{2}}[/tex]

    I thought that it's the value stated in the question, but that seemed too easy, but this other method gives messed up units. Maybe there's another way, or I was correct first time.

    2. I'm not sure how to do this at all :frown: so some advice would be great.

    3. Again, not sure how to do this. Sure it's not a tricky calculation though.

    Perhaps conservation of energy and conservation of momentum are playing their part in these 2 questions.

    4. Expression for momentum is [itex]P=mv[/itex] so I just need to rearrange the equation, and find out the velocity of the tauon. Not sure how to calculate this. Unless they are moving at [itex]v=c[/itex], that's just a thought.

    .. SO I don't think I'm too far from the answers, hopefully getting the right ideas, just could do with some advice on the tricky bits. :smile:
  2. jcsd
  3. Apr 28, 2010 #2
    1) Not sure, I would have thought 5000 MeV as well. Or do they want just the energy from the rest mass of the electron. Makes no sense really. Might have a definition in your book for this.

    2) Use conservation of momentum

    3) This is just conservation of energy, you won't need conservation of momentum for this.

    4) I believe they want rest mass of the tauon. Also, don't use the classical equation for momentum, there will be a [itex]\gamma[/itex] involved. And you will need your answer from (3) to help you.
  4. Apr 29, 2010 #3
    1) I'm glad I'm not the only one who thinks that the wording of this part of the question isn't very clear. It's rather confusing.

    Looking up 'mass energy' on Wikipedia:


    I think that this appears to mean that what is being asked for is:

    [tex]E=mc^{2}=m_{e}c^{2}=(9.11\times 10 ^{-31}kg)(3\times 10^{8})=8.2\times 10^{-14}kgms^{-1}=8.2\times 10^{-14}J[/tex]

    I do hope it's that, I have no other ideas what it might be and that seems most likely.

    2) Conservation of momentum:


    So if mass is conserved, can apply this to the collision in question, hence:


    So this shows that [itex]\tau^{-}[/itex] and [itex]\tau^{+}[/itex] have equal magnitudes of momentum.

    But how do I show that they move in exactly opposite directions?

    3) Conservation of energy: [itex]E_{initial}=E_{final}[/itex]

    Hence: [tex]E_{initial}=2(5000MeV)=10000MeV=E_{final}[/tex]

    So: [tex]E_{\tau^{-}}=E_{\tau^{+}}=\frac{10000MeV}{2}=5000MeV[/tex]

    So basically the masses of the [itex]\tau^{-}, \tau^{+}, e^{-}, e^{+}[/itex] are all the same? I guess this is right, through conservation, just seems a bit odd though.

    4) From part 3): [itex]E_{\tau}=5000MeV[/itex]

    The momentum given in the question: [itex]P_{\tau}=4675MeV/c[/itex]

    Don't use classical equation for momentum [itex]P=mv[/itex] since relativistic correction of Lorentz factor [itex]\gamma[/itex] would need to be considered.

    So I know [itex]E_{\tau}[/itex] and [itex]P_{\tau}[/itex] and need to find [itex]m_{\tau}[/itex].

    I thought I could just use [itex]E=mc^{2}\implies m=\frac{E}{c^{2}}[/itex] but this is the same method as for part 1) and would give the same answer, and I don't see
    how the given momentum factors in and why it would be there if it wasn't meant to be used.

    .. Hopefully I'm getting somewhere with all this now :smile:
  5. Apr 29, 2010 #4
    (2) What is the total momentum before the collision?

    (3) The rest masses are not the same. Just the energy of the individual particles.

    (4) Again you are confusing rest mass with relativistic mass. They want to know the rest mass of the tau particle. And remember:

    [tex]E=\gamma m c^2[/tex]

    So you have 2 unknowns, [itex]v[/itex] and [itex]m[/itex]. So you will use the 2 equations you have, [itex]p[/itex] and [itex]E[/itex], to solve for the 2 unknowns.
  6. Apr 30, 2010 #5

    Total momentum before the collision is:


    But I don't know what [itex]v_{elec}[/itex] and [itex]v_{posi}[/itex], and would have to look up [itex]m_{elec}[/itex] and [itex]m_{posi}[/itex]

    It feels that I'm obviously missing something really straighforward here. That's annoying.

    Also not sure if that equation for momentum should have a [itex]\gamma[/itex] variable. I guess it probably should assuming electron and positron are both travelling a speed that's a significant proportion of light speed.


    Ok, so is it correct to say, as in my previous post, that:



    So just not with the following comment that the masses are the same, as that's untrue.


    From part 3) I now know that: [itex]m_{\tau}=5000MeV[/itex]

    The momentum is given as: [itex]|P|=4675Mev/c[/itex]

    I need to find the rest mass of the tau particle.

    Firstly I don't see why I can't use this:


    I thought I could do this using this equation:


    Which was found here:


    So that gives the result:

    [tex]m_{0}=\sqrt{(5000)^{2}-(4675)^{2}}=3.14\times 10^{6}(kg?)[/tex]

    Which is obviously a rediculously large answer for the rest mass of the tau particle. This must be therefore incorrect.

    So then I thought maybe instead I use this equation:

    [tex]E=\gamma m c^{2}\implies m=\frac{E}{\gamma c^{2}}[/tex]

    Which should work if I knew the value of [itex]v[/itex]. Presuming this isn't the equation to give the relativistic mass instead, which it could well be.

    I could use:

    [tex]P=mv\implies v=\frac{P}{m}[/tex]

    But I don't know what [itex]m[/itex] is, that's what I'm trying to find.

    SO basically I'm now stuck on how I use some of the above in order to presumably find [itex]v[/itex] and then therefore be able to find the rest mass [itex]m[/itex]

    Perhaps I can combine the equations for energy and momentum:

    [tex]E=m\gamma c^{2}[/tex] and [tex]P=mv[/tex]

    To give:


    Then use this value to calculate rest mass:


    How about all that?
  7. Apr 30, 2010 #6
    (2) They have the exact same energy, so their momentum must be the same (since they have the same rest masses). Since they collide head on, then their momentum vectors must be opposite of each other. (Note: These are relativistic speeds so you need the [itex]\gamma[/itex])

    (3) Yes, due to conservation of energy the individual energies of the tauons should be the same as the individual energies of the electrons.

    (4) You must use relativistic energy and momentum and not classical momentum. So you will see [itex]\gamma[/itex] whenever you see [itex]m[/itex].

    Also, you can use the equation with energy and momentum to find the invariant mass that you got from wikipedia, but wikipedia also sets [itex]c=1[/itex] and so you have to be extremely careful with the units. In your case, you are working with MeV and for some reason you assign the units of the mass as kg.

    The way to write that equation with [itex]c[/itex] is:

    [tex]m_0 c^2 = \sqrt{E^2-p^2 c^2}[/tex]

    If you use units of MeV for [itex]E[/itex], then the units of [itex]m_0 c^2[/itex] will also be MeV.
  8. Apr 30, 2010 #7

    So the equation should be this:

    [tex]P=mv=\left(m_{elec}\gamma v_{elec}\right)-\left(m_{posi}\gamma v_{posi}\right)[/tex]


    But then I still don't know how to calculate the velocities?


    So to calculate the rest mass of the tauon, use the formula:

    [tex]m_{0}c^{2} = \sqrt{E^{2}-p^{2}c^{2}}[/tex]

    Where [itex]E=5000MeV[/itex] and [itex]P=4675MeV[/itex]

    Then do I use [itex]c=1[/itex]? since using [itex]c=3\times 10^{8}[/itex] gives the square root of a negative number which is an error.
  9. Apr 30, 2010 #8
    (2) They have the same energy and masses (also they are heading toward each other), so their momentums are equal but opposite. When you add two vectors of equal magnitude and opposite directions, what do you get?

    (4) The units of "p" are MeV/c, please re-read your first post.

    EDIT: Corrected my units of momentum.
    Last edited: Apr 30, 2010
  10. Apr 30, 2010 #9
    .. 0? :redface:

    Indeed they are :smile:

    So the [itex]c^{2}[/itex] units terms cancel if that makes sense, and equation should read:

    [tex]m_{0}c^{2} = \sqrt{E^{2}-p^{2}c^{2}}=\sqrt{5000^{2}-4675^{2}}=1.77\times 10^{3}MeV[/tex]

    .. is this getting any better now?
  11. Apr 30, 2010 #10
    (2) That is the initial total momentum, what can you deduce about the final momentum from this using conservation of momentum.

    (4) Yes that is correct. Now solve for mass. I don't know if they prefer the units in MeV/c^2 or kg.
  12. May 1, 2010 #11
    Ok, great.

    So then:


    Momentum before collision (or the initial momentum) is 0, so from conservation of momentum, the momentum after the collision (or the final momentum) is 0.

    Going back to the question:

    So both the tau particles have equal magnitudes of momentum of 0. Right?

    How do I show they move in exactly opposite directions?


    [tex]m_{0}c^{2} = \sqrt{E^{2}-p^{2}c^{2}}=\sqrt{5000^{2}-4675^{2}}=1.77\times 10^{3}MeV[/tex]

    this is the correct units yes? (i.e. just MeV, there's no c term here)

    Then the mass:

    [tex]m_{0}=\frac{1.77\times 10^{3}MeV}{c^{2}}=1.97\times 10^{-14}MeV/c^{2}[/tex]

    Then parts (1) and (4) have already been done.
  13. May 1, 2010 #12
    (2) Their momentum are not 0. Just the total final momentum is 0.

    (4) What happened in the last equation? Looks like you divided out c^2, but also left it there in the units. You can convert MeV/c^2 to kg if you like. You can find the conversion online. Or just leave it is MeV/c^2, but do not divide out the c^2, just leave it in the units.
  14. May 1, 2010 #13
    (2) Right so the total momentum after the collision is 0, just like it is 0 before the collision. Therefore the momentum of the [itex]\tau[/itex] particles, + and -, must be the same in order for the result to be 0, but the momentums themselves are not the same.

    I.e. using a really basic example:

    [tex]P[t+]=5, P[t-]=-5, so |P[t+]|-|P[t-]|=5-5=0[/tex]

    Hopefully that makes sense I know it's rather crude! :wink:

    (4) I think I just got a little bit confused. I have this:

    [tex]m_{0}c^{2} = \sqrt{E^{2}-p^{2}c^{2}}=\sqrt{5000^{2}-4675^{2}}=1.77\times 10^{3}MeV[/tex]

    Which I beleive to be correct, and with correct units of [itex]MeV[/itex].

    Then I thought the rest mass should be in units of [itex]MeV/c^{2}[/itex] so I divided through the result by [itex]c^{2}[/itex]. That's why those units of [itex]MeV/c^{2}[/itex] are there.

    [tex]m_{0}=\frac{1.77\times 10^{3}MeV}{c^{2}}=1.97\times 10^{-14}MeV/c^{2}[/tex]

    I get a bit confused where [itex]Mev[/itex]'s and [itex]c^{2}[/itex]'s should be or not be :uhh:
  15. May 1, 2010 #14
    (2) Just use vectors:



    So they are equal and opposite.

    (4) If you list c^2 in the units then you can't divide it from your number. Otherwise you are dividing it twice.
  16. May 1, 2010 #15
    (2) So I wouldn't have to put [itex]||[/itex] notation in there?!

    (4) Ok so it should be:

    [tex]m_{0}=\frac{1.77\times 10^{3}MeV}{c^{2}}=1.97\times 10^{-14}MeV}[/tex]
  17. May 1, 2010 #16
    You would be wrong again. Because by writing 'c' out in its full form, it also has units of 'm/s'. Where did those units go? It is customary to not expand 'c' out but just leave it as MeV/c^2. You are just doing more unnecessary work.
  18. May 1, 2010 #17
    .. so just leave it as this:

    m_{0}=1.97\times 10^{-14}MeV/c^{2}

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